The theorem is properly applied:

' "cos" x = 0` when `x= ... -(3 pi)/(2), -(pi)/(2), (pi)/(2), (3 pi)/(2), ...`
and `"cos" x ` is continuous and differentiable everywhere.

Thus `"sec" x` is continuous and differentiable on any interval not containing one of
`... -(3 pi)/(2), -(pi)/(2), (pi)/(2), (3 pi)/(2), ...`

The interval `[-(pi)/(6) , (pi)/(3) ]` does not contain any of these, so ` "sec" x` is differentiable on `(-(pi)/(6) , (pi)/(3) )` and continuous on `[-(pi)/(6) , (pi)/(3) ]` and we may apply the mean value theorem.

We consider the endpoints of the interval:

`"sec" -(pi)/(6) = (2sqrt(3))/(3)`

`"sec" (pi)/(3) = 2`

If we were to draw a line through the two points

`(-(pi)/(6) , (2sqrt(3))/(3) )` and `( (pi)/(3) , 2 )`

it would have slope:

`(2 - (2sqrt(3))/(3) )/((pi)/(3)+ (pi)/(6) ) `

or

`((6 - 2sqrt(3))/(3))/((pi)/(2)) = (12 - 4sqrt(3))/(3 pi)`


So, by the mean value theorem, there is some point c in the interval `[-(pi)/(6) , (pi)/(3) ]` such that

`f'(c) = (12 - 4sqrt(3))/(3 pi)`

In pictures:

The black line is the function

The blue line is the line you get by connecting the endpoints of the interval

The dashed blue line has the same slope as the blue line, but is tangent to the graph. (roughly, I just eyeballed it)

This is the crux of the mean value theorem:

Somewhere in that interval, the tangent line has the same slope as the line you get by connecting the endpoints